Question

Differentiate the following seven functions, applying the process of \(\$ 1\) step by step. \(f(x)=1-2 x^{4}\)

Short answer

Answer: The derivative of the function \(f(x) = 1 - 2x^4\) with respect to \(x\) is \(f'(x) = 8x^3\).

Step-by-step solution

Identify the components of the function

We first need to identify the different components of the function \(f(x) = 1 - 2x^4\). In this function, we have a constant term, \(1\), and a term involving a power of \(x\), \(-2x^4\).

Apply the constant rule to the constant term

Applying the constant rule to the constant term \(1\), we find that its derivative with respect to \(x\) is zero. In other words, \(\frac{d(1)}{dx} = 0\).

Apply the power rule to the term involving a power of x

Now we need to apply the power rule to the term \(-2x^4\). According to the power rule, the derivative of \(x^n\) is \(nx^{n-1}\). In this case, \(n = 4\), so we have \(\frac{d(-2x^4)}{dx} = -2 (4x^{4-1}) = -8x^3\).

Apply the sum and difference rule

Now that we have the derivatives of the individual terms, we can apply the sum and difference rule to find the derivative of the entire function \(f(x)\). According to the sum and difference rule, the derivative of a sum/difference is the sum/difference of their derivatives. So, the derivative of \(f(x) = 1 - 2x^4\) is given by: \(f'(x) = \frac{d(1)}{dx} - \frac{d(2x^4)}{dx} = 0 - (-8x^3) = 8x^3\).

Write down the final answer

After differentiating the function step by step, we've found that the derivative of \(f(x) = 1 - 2x^4\) is \(f'(x) = 8x^3\).

Key concepts

Constant Rule

The constant rule is one of the simplest rules in differentiation. It is a special rule that applies when you need to find the derivative of a constant term, like the number 1 in our function.
Regardless of the value of the constant, its derivative is always zero. This is because a constant does not change, and differentiation measures how a function changes.
To put it clearly, if you have a constant function, say \(c\), the derivative \(\frac{d(c)}{dx} = 0\).
In our exercise, this is seen when finding the derivative of the term \(1\), where we apply this rule and get \( \frac{d(1)}{dx} = 0 \).
The use of the constant rule simplifies the differentiation process by effectively removing any constant terms. Without constants contributing to changes in values, this is an intuitive step. It allows for focus on variable terms that actually change with respect to \(x\).

Power Rule

The power rule is a fundamental technique in calculus for differentiating expressions of the form \(x^n\). When you have an expression like this, the power rule helps transform it by bringing down the exponent as a coefficient and then reducing the exponent by one.
To use the power rule, consider the expression \( x^n \). Its derivative is computed by the formula \( nx^{n-1} \).
For example, in our exercise, we consider \(-2x^4\). Here, \(n = 4\), so the derivative is calculated as:\(-2 \cdot 4x^{4-1} = -8x^3\).
This method is extremely efficient for polynomials and simplifies many derivative calculations.
Remember, the power rule only applies to powers of \(x\), not other functions. In cases that involve coefficients, as seen in \(-2x^4\), simply multiply the calculated derivative by the coefficient. This is a versatile tool in any calculus student's toolkit, making seemingly complex operations much easier.

Sum and Difference Rule

The sum and difference rule is useful when you have a function that involves adding or subtracting two or more terms. The beauty of this rule is that you can separate each term, find its derivative individually, and then combine them back together as a sum or difference.
This means if you have a function like \(f(x) = u(x) + v(x)\) or \(f(x) = u(x) - v(x)\), the derivative \(f'(x)\) is simply \(u'(x) + v'(x)\) or \(u'(x) - v'(x)\) respectively.
In the problem provided, the sum and difference rule helps to calculate the derivative of \(f(x) = 1 - 2x^4\). After finding the derivatives of each term: zero for the constant and \(-8x^3\) for the variable term, we subtract accordingly:\(f'(x) = 0 - (-8x^3) = 8x^3\).
This rule reduces potential errors and streamlines the solving process. It's essentially about keeping things separate and neat, making differentiation tasks more manageable. The sum and difference rule is indispensable for working with more complex polynomial functions, aiding in precise derivative calculations.